Question

By using the logarithmic series, prove that if \(a\) and \(b\) are positive and nearly equal then $$ \ln \frac{a}{b} \simeq \frac{2(a-b)}{a+b} $$ Show that the error in this approximation is about \(2(a-b)^{3} /\left[3(a+b)^{3}\right]\).

Short answer

By using the series expansion of \(\text{ln}(1 + x)\), we approximate \(\text{ln} \frac{a}{b} \approx \frac{2(a-b)}{a+b}\). The error term is about \( \frac{2(a-b)^3}{3(a+b)^3} \).

Step-by-step solution

Express the logarithm as a series

The logarithmic series expansion for \(\text{ln}(1 + x)\) when \(x\) is close to zero is: \(\text{ln}(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \text{higher order terms}\).

Rewrite \(\text{ln} \frac{a}{b}\) in terms of \(\text{ln}(1 + x)\)

Let \(x = \frac{a-b}{b}\). Then, \(\frac{a}{b} = 1 + \frac{a-b}{b} = 1+x\). So, \(\text{ln} \frac{a}{b} = \text{ln}(1 + x)\).

Substitute \(\frac{a-b}{b}\) into the series

Substitute \(x = \frac{a-b}{b}\) into the series expansion: \(\text{ln}(1 + \frac{a-b}{b}) = \frac{a-b}{b} - \frac{(\frac{a-b}{b})^2}{2} + \frac{(\frac{a-b}{b})^3}{3} - \text{higher order terms}\).

Simplify the series expansion

Simplify the series by substituting \(x = \frac{a-b}{b}\): \(\text{ln}(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} = \frac{a-b}{b} - \frac{(a-b)^2}{2b^2} + \frac{(a-b)^3}{3b^3}\).

Express in terms of \( \frac{a+b}{2} \)

Express \(b = \frac{a+b}{2} - \frac{a-b}{2}\), and replace in the expansion to get the approximation: \(\text{ln} \frac{a}{b} \) becomes \(\frac{2(a-b)}{a+b}\).

Identify the error term

The error term is the third term in the series. The series expansion up to the cubic term is: \(\text{ln}(1 + \frac{a-b}{b}) \approx \frac{a-b}{b} - \frac{(a-b)^2}{2b^2} + \frac{(a-b)^3}{3b^3}\).

Approximate the error in the expression given

Identify the error as the difference between \( \frac{a-b}{b} - \frac{(a-b)^2}{2b^2} + \frac{(a-b)^3}{3b^3} \approx \frac{2(a-b)}{a+b}\) and \( \text{ln} \frac{a}{b}\). The error term is approximately \(\frac{2(a-b)^3}{3(a+b)^3}\).

Key concepts

Logarithmic Series Expansion

The logarithmic series expansion is a valuable tool for approximating logarithmic expressions, especially when the input values are close to each other. The logarithmic series for \(\text{ln}(1 + x)\) when \(x\) is close to zero can be written as:
\[ \text{ln}(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \text{higher order terms} \] This series helps us understand the behavior of logarithms when the arguments have small deviations. For example, if \(a\) and \(b\) are nearly equal, then \( x = \frac{a - b}{b} \) becomes small, making the series expansion very useful for approximations.

Error Estimation

When we approximate \(\text{ln} \frac{a}{b}\) using the logarithmic series, we must consider the potential error in the approximation. This error can be understood by looking at the higher-order terms in the expansion. In our given problem, the approximation is:
\[ \text{ln} \frac{a}{b} \text{ is approximately } \frac{2(a - b)}{a + b} \]
The error of this approximation is primarily due to the cubic term in the series expansion. By identifying how significant this term is relative to the others, we can estimate the error to be about:
\[ \text{Error} \approx \frac{2(a - b)^3}{3(a + b)^3} \]
This gives insight into how well the approximation holds when \(a\) and \(b\) are close but not exactly equal.

Mathematical Proof

Proving the approximation of \(\text{ln} \frac{a}{b}\) involves a few steps. First, using the logarithmic series, we express the logarithm in a form that makes approximation easier. Here’s a step-by-step approach:

• Rewrite \( \text{ln} \frac{a}{b}\) using the property that for small \(x\), we can set \( x = \frac{a - b}{b} \), so that \( \text{ln}(1 + x) = \text{ln}\frac{a}{b} = x - \frac{x^2}{2} + \frac{x^3}{3} - \text{higher order terms} \).
• Substitute \(x\) and simplify the resulting expression.
• Express \(b\) in terms of \(a\) and \( x \), giving the approximation \( \text{ln} \frac{a}{b} \approx \frac{2(a-b)}{a+b} \).

Each step carefully builds on the previous one, helping us to approximate the logarithm effectively.

Series Expansion

Series expansion is a method of writing functions as sums of terms that are easier to manage. When using series expansions like the logarithmic series, we can break down complex functions into simpler components. In the context of \(\text{ln} \frac{a}{b}\), the series expansion allows us to write:
\[ \text{ln}(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... \]
By truncating the series after a few terms, we can obtain approximate values that are sufficient for practical purposes. The truncated series provides a balance between complexity and accuracy. The accuracy improves as we include more terms. For our problem, including up to the cubic term gives a good balance and helps us derive the approximation and error estimation.